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Simora [160]
3 years ago
8

Julie tried to use the law of syllogism to draw a conclusion based on the statements below. Explain why she is not able to do so

. If AB is bisected by another line segment, two pairs of congruent angles are formed. If AB is bisected by another line segment, segment AB will be cut into two equal sections.
Mathematics
2 answers:
posledela3 years ago
7 0

Answer:

Syllogism states: If p⇒q, and q⇒r, then p⇒r.

In the given case, there is no implication q⇒r.

But she is using,

p⇒q

p⇒r (directly, without the proposition q⇒r).

Therefore, she cannot draw a conclusion based on the statements.

Step-by-step explanation:

The law of syllogism starts from two different premises to reach a conclusion; This law consists of:

Premise 1: If p implies q; and

Premise 2: q implies r; that equals

Conclusion: p implies r.

If p⇒q, and q⇒r, then p⇒r.

In the given case, we define the propositions like this:

p = AB is crossed by another line segment

q = two pairs of congruent angles are formed

r = segment AB will be cut into two equal sections

We can see that the syllogism law is not being applied, because there is no implication q⇒r.

The following operations are being presented:

p⇒q

p⇒r (directly, without the proposition q⇒r).

Therefore, she cannot draw a conclusion based on the statements.

Hope this helps!

ira [324]3 years ago
4 0

Answer:  The conditional statements are not in the correct form to make a conclusion using the law of syllogism. “If p, then q and if p, then r” cannot be used to draw a conclusion using the law of syllogism. The law of syllogism could be used if the hypothesis in the second statement was "if two pairs of congruent angles are formed."

Step-by-step explanation:

"If p, then q and if p, then r" cannot be used to draw a conclusion using the law of syllogism.

Neither of the conclusions of the conditional statements are the hypothesis of the other.

"If two pairs of congruent angles are formed" could be the hypothesis of the second statement.

** Both can be used to answer the question :)

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Step-by-step explanation:

\large\underline{\sf{Solution-}}

Given expression is

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

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Now, Consider Denominator, we have

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {n}^{n}

can be rewritten as

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {(n - 1)}^{n} +   {n}^{n}

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\rm \:  =  \:  {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]

Now, Consider

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

So, on substituting the values evaluated above, we get

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{\dfrac{ {n}^{n}  - 1}{1 -  \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

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